Theorem ndmovg 7600

Description: The value of an operation outside its domain. (Contributed by NM, 28-Mar-2008.)

Assertion

Ref	Expression
ndmovg	⊢ ((dom 𝐹 = (𝑅 × 𝑆) ∧ ¬ (𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑆)) → (𝐴𝐹𝐵) = ∅)

Proof of Theorem ndmovg

Step	Hyp	Ref	Expression
1		df-ov 7419	. 2 ⊢ (𝐴𝐹𝐵) = (𝐹‘⟨𝐴, 𝐵⟩)
2		eleq2 2851	. . . . . 6 ⊢ (dom 𝐹 = (𝑅 × 𝑆) → (⟨𝐴, 𝐵⟩ ∈ dom 𝐹 ↔ ⟨𝐴, 𝐵⟩ ∈ (𝑅 × 𝑆)))
3		opelxp 5695	. . . . . 6 ⊢ (⟨𝐴, 𝐵⟩ ∈ (𝑅 × 𝑆) ↔ (𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑆))
4	2, 3	bitrdi 290	. . . . 5 ⊢ (dom 𝐹 = (𝑅 × 𝑆) → (⟨𝐴, 𝐵⟩ ∈ dom 𝐹 ↔ (𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑆)))
5	4	notbid 321	. . . 4 ⊢ (dom 𝐹 = (𝑅 × 𝑆) → (¬ ⟨𝐴, 𝐵⟩ ∈ dom 𝐹 ↔ ¬ (𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑆)))
6		ndmfv 6914	. . . 4 ⊢ (¬ ⟨𝐴, 𝐵⟩ ∈ dom 𝐹 → (𝐹‘⟨𝐴, 𝐵⟩) = ∅)
7	5, 6	biimtrrdi 257	. . 3 ⊢ (dom 𝐹 = (𝑅 × 𝑆) → (¬ (𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑆) → (𝐹‘⟨𝐴, 𝐵⟩) = ∅))
8	7	imp 412	. 2 ⊢ ((dom 𝐹 = (𝑅 × 𝑆) ∧ ¬ (𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑆)) → (𝐹‘⟨𝐴, 𝐵⟩) = ∅)
9	1, 8	eqtrid 2809	1 ⊢ ((dom 𝐹 = (𝑅 × 𝑆) ∧ ¬ (𝐴 ∈ 𝑅 ∧ 𝐵 ∈ 𝑆)) → (𝐴𝐹𝐵) = ∅)

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 401   = wceq 1570   ∈ wcel 2145  ∅c0 4282  ⟨cop 4593   × cxp 5657  dom cdm 5659  ‘cfv 6537  (class class class)co 7416

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2734  ax-sep 5255  ax-nul 5267  ax-pr 5402

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-mo 2566  df-eu 2596  df-clab 2741  df-cleq 2754  df-clel 2837  df-ne 2958  df-ral 3079  df-rex 3089  df-rab 3415  df-v 3455  df-dif 3905  df-un 3907  df-in 3909  df-ss 3919  df-nul 4283  df-if 4486  df-sn 4588  df-pr 4590  df-op 4594  df-uni 4871  df-br 5108  df-opab 5172  df-xp 5665  df-dm 5669  df-iota 6493  df-fv 6545  df-ov 7419

This theorem is used by:  ndmov  7601  curry1val  8105  curry2val  8109  1div0  11900  repsundef  14844  cshnz  14865  mamufacex  22619  mavmulsolcl  22774  mavmul0g  22776  iscau2  25506  1div0apr  30934  rrxsphere  49665